What is the integral of ((x^3+2x))/(sqrt(x^2+4)) ?

The integral of ((x^3+2x))/(sqrt(x^2+4)) is 1/3 (x^2+4)^{3/2}-2sqrt(x^2+4)+C

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integral of ((x^3+2x))/(sqrt(x^2+4))

\int \frac{( x ^{3} + 2 x )}{x ^{2} + 4} d x

\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4 + C

Solution steps

\int \frac{x ^{3} + 2 x}{x ^{2} + 4} d x

Expand

\frac{x ^{3} + 2 x}{x ^{2} + 4} : \frac{x ^{3}}{x ^{2} + 4} + \frac{2 x}{x ^{2} + 4}

= \int \frac{x ^{3}}{x ^{2} + 4} + \frac{2 x}{x ^{2} + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{3}}{x ^{2} + 4} d x + \int \frac{2 x}{x ^{2} + 4} d x

\int \frac{x ^{3}}{x ^{2} + 4} d x = \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4

\int \frac{2 x}{x ^{2} + 4} d x = 2 x^{2} + 4

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4 + 2 x^{2} + 4

Simplify

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4

Add a constant to the solution

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4 + C

What is the integral of ((x^3+2x))/(sqrt(x^2+4)) ?
The integral of ((x^3+2x))/(sqrt(x^2+4)) is 1/3 (x^2+4)^{3/2}-2sqrt(x^2+4)+C